Calculate the pH after 12.50 mL of 0.100 M NaOH has been added to 25.0 mL

Question

Calculate the pH after 12.50 mL of 0.100 M NaOH has been added to 25.0 mL of 0.100 M CH3CO2H. The Ka of CH3CO2H is 1.8×10−5.

Answer 1

Hello Aisha

This question illustrates an acid-base neutralization reaction, specifically the reaction of a weak acid (CH3CO2H) with a strong base (NaOH). The neutralization reaction forms salt and water as the products as indicated below:

CH3CO2H (aq) + NaOH (aq) --> CH3CO2Na (aq) + H2O (l)

Na is a spectator ion, so we can omit it from the above reaction to give the net ionic equation for this neutralization reaction.

CH3CO2H (aq) + OH^- (aq) --> CH3CO2^- (aq) + H2O (l)

Using the given Molarity and volume of NaOH and CH3CO2H, calculate the # of moles of each substance.

mols of NaOH reacted = 0.100 mol/L x 12.50 mL x (1 L / 1000 mL) = 0.00125 mol NaOH

mol of CH3CO2H neutralized = 0.100 mol/L x 25.0 mL x (1 L / 1000 mL) = 0.00250 mol CH3CO2H

From the above values, you can see that NaOH is the limiting reagent. 0.00125 moles of NaOH will react completely to form CH3CO2^- after it neutralizes the acid (CH3CO2H). Since CH3CO2H is in excess, let's calculate how much CH3CO2H is remaining.

mol CH3CO2H remaining = 0.00250 - 0.00125 = 0.00125 mol CH3CO2H

mol CH3CO2^- formed = 0.00125 mol

Now calculate the concentration of CH3CO2H and CH3CO2^- present in the solution.

Final volume of solution after neutralization = 12.50 + 25.0 = 37.5 mL x (1 L / 1000 mL) = 0.0375 L

Molarity of CH3CO2H = 0.0125 mol / 0.0375 L = 0.333 M

Molarity of CH3CO2^- = 0.0125 mol / 0.0375 L = 0.333 M

To calculate the pH, you can use the Hendersen-Haaselbach equation.

pH = pKa + log ([CH3CO2^-] / [CH3CO2H])

Ka = 1.8 x 10-5

pKa = - log Ka = - log 1.8 x 10-5 = 4.74

pH = 4.74 + log (0.333 / 0.333) = 4.74 + log 1 = 4.74 + 0 = 4.74